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EE 5340Semiconductor Device TheoryLecture 4 - Fall 2003
Professor Ronald L. [email protected]
http://www.uta.edu/ronc
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Web Pages
* If you have not already done so:• R. L. Carter’s web page
– www.uta.edu/ronc/
• EE 5340 web page and syllabus– www.uta.edu/ronc/5340/syllabus.htm
• University and College Ethics Policies– www2.uta.edu/discipline/– www.uta.edu/ronc/5340/COE_EthicsStatement_Fall02.htm
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Assignment 1: If you have not received response from me ...• Send e-mail to [email protected]
– On the subject line, put “5340 e-mail”– In the body of message include
• email address: ______________________• Your Name*: _______________________• Last four digits of your Student ID: _____
* Your name as it appears in the UTA Record - no more, no less
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Assignment 2: 930AM stu-dents only*• IF you have a class OTHER than 5340
at 8 AM, e-mail to [email protected]– Subject line: “5340 8 AM Class”– In the body of message include
• email address: ______________________• Your Enrollment Name*: ______________• Class you are taking at 8:00 AM: ________
* If you don’t do this, and don’t take Test 1 at 8:00 AM, you will not get credit for Test 1
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Classes ofsemiconductors• Intrinsic: no = po = ni, since Na&Nd << ni
=[NcNvexp(-Eg/kT)]1/2, (not easy to get)
• n-type: no > po, since Nd > Na, noNd-Na
• p-type: no < po, since Nd < Na, poNa-Nd
• Compensated: no=po=ni, w/ Na- = Nd
+ > 0
• Note: n-type and p-type are usually partially compensated since there are usually some opposite- type dopants
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Equilibriumconcentrations• Charge neutrality requires
q(po + Nd+) + (-q)(no + Na
-) = 0
• Assuming complete ionization, so Nd
+ = Nd and Na- = Na
• Gives two equations to be solved simultaneously
1. Mass action, no po = ni2, and
2. Neutrality po + Nd = no + Na
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Equilibriumconc (cont.)• For Nd > Na (taking the + root) no
= (Nd-Na)/2 + {[(Nd-Na)/2]2+ni2}1/2
• For Nd >> Na and Nd >> ni, can use the binomial expansion, giving
no = Nd/2 + Nd/2[1 + 2ni2/Nd
2 + … ]
• So no = Nd, and po = ni2/Nd in the limit
of Nd >> Na and Nd >> ni
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Equilibriumconc (cont.)• For Na > Nd (taking the + root) po
= (Na-Nd)/2 + {[(Na-Nd)/2]2+ni2}1/2
• For Na >> Nd and Na >> ni, can use the binomial expansion, giving
po = Na/2 + Na/2[1 + 2ni2/Na
2 + … ]
• So po = Na in the limit of Na >> Nd and Na >> ni
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Examplecalculations• For Nd = 3.2E16/cm3, ni = 1.4E10/cm3
no = Nd = 3.2E16/cm3
po = ni2/Nd , (po is always ni
2/no)
= (1.4E10/cm3)2/3.2E16/cm3
= 6.125E3/cm3 (comp to ~1E23 Si)
• For po = Na = 4E17/cm3,
no = ni2/Na = (1.4E10/cm3)2/4E17/cm3
= 490/cm3
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Position of theFermi Level• Efi is the Fermi level
when no = po
• Ef shown is a Fermi level for no > po
• Ef < Efi when no < po
• Efi < (Ec + Ev)/2, which is the mid-band
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EF relative to Ec and Ev• Inverting no = Nc exp[-(Ec-EF)/kT] gives
Ec - EF = kT ln(Nc/no) For n-type material: Ec - EF =kTln(Nc/Nd)=kTln[(Ncpo)/ni
2]
• Inverting po = Nv exp[-(EF-Ev)/kT] gives EF - Ev = kT ln(Nv/po) For p-type material: EF - Ev = kT ln(Nv/Na)
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EF relative to Efi
• Letting ni = no gives Ef = Efi
ni = Nc exp[-(Ec-Efi)/kT], soEc - Efi = kT ln(Nc/ni). ThusEF - Efi = kT ln(no/ni) and for n-typeEF - Efi = kT ln(Nd/ni)
• Likewise Efi - EF = kT ln(po/ni) and for p-type Efi - EF = kT ln(Na/ni)
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Locating Efi in the bandgap • Since Ec - Efi = kT ln(Nc/ni),
and Efi - Ev = kT ln(Nv/ni)
• The 1st equation minus the 2nd givesEfi = (Ec + Ev)/2 - (kT/2) ln(Nc/Nv)
• Since Nc = 2.8E19cm-3 > 1.04E19cm-3 = Nv, the intrinsic Fermi level lies below the middle of the band gap
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Samplecalculations• Efi = (Ec + Ev)/2 - (kT/2) ln(Nc/Nv), so at
300K, kT = 25.86 meV and Nc/Nv = 2.8/1.04, Efi is 12.8 meV or 1.1% below mid-band
• For Nd = 3E17cm-3, given that Ec - EF = kT ln(Nc/Nd), we have Ec - EF
= 25.86 meV ln(280/3), Ec - EF = 0.117 eV =117meV ~3x(Ec - ED) what Nd
gives Ec-EF =Ec/3
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Equilibrium electronconc. and energies
o
v2i
vof
i
ofif
fif
i
o
c
ocf
cf
c
o
pN
lnkTn
NnlnkTEvE and
;nn
lnkTEE or ,kT
EEexp
nn
;Nn
lnkTEE or ,kT
EEexp
Nn
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Equilibrium hole conc. and energies
o
c2i
cofc
i
offi
ffi
i
o
v
ofv
fv
v
o
nN
lnkTn
NplnkTEE and
;np
lnkTEE or ,kT
EEexp
np
;Np
lnkTEE or ,kT
EEexp
Np
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Carrier Mobility
• In an electric field, Ex, the velocity (since ax = Fx/m* = qEx/m*) is
vx = axt = (qEx/m*)t, and the displ
x = (qEx/m*)t2/2
• If every coll, a collision occurs which “resets” the velocity to <vx(coll)> = 0, then <vx> = qExcoll/m* = Ex
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Carrier mobility (cont.)• The response function is the
mobility.• The mean time between collisions,
coll, may has several important causal events: Thermal vibrations, donor- or acceptor-like traps and lattice imperfections to name a few.
• Hence thermal = qthermal/m*, etc.
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Carrier mobility (cont.)• If the rate of a single contribution
to the scattering is 1/i, then the total scattering rate, 1/coll is
all
collisions itotal
all
collisions icoll
11
by given is mobility total
the and , 11
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Drift Current
• The drift current density (amp/cm2) is given by the point form of Ohm LawJ = (nqn+pqp)(Exi+ Eyj+ Ezk), so
J = (n + p)E = E, where
= nqn+pqp defines the conductivity
• The net current is SdJI
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Net silicon extrresistivity (cont.)• Since = (nqn + pqp)-1, and
n > p, ( = q/m*) we have
p > n
• Note that since1.6(high conc.) < p/n < 3(low conc.), so
1.6(high conc.) < n/p < 3(low conc.)